Respuesta :
The equation of plane containing the points ABC is 13x+y-18z = 1035.
What is equation of plane?
A plane in 3D coordinate space is determined by a point and a vector that is perpendicular to the plane. The equation of plane is defined as the position vectors and the Cartesian product of the vector perpendicular to the plane.
For the given situation,
The plane containing the points
A(10, 5, -50), B(15, 30, -45) and C(-5, 20, -60).
The equation of the plane in the form r.n=d can be represented as
[tex](r-a).[(b-a)[/tex] × [tex](c-a)]=0[/tex] ------ (1)
Let r = xi + yj +zk
a = 10i + 5j - 50k
b = 15i + 30j - 45k
c = -5i + 20j - 60k
[tex](r-a)=(x-10)i+(y-5)j+(z+50)k[/tex] ------- (2)
[tex](b-a)=(15-10)i+(30-5)j+(-45-(50))k[/tex]
⇒ [tex](b-a)=5i+25j+5k[/tex]
[tex](c-a)=(-5-10)i+(20-5)j+(-60-(50))k[/tex]
⇒ [tex](c-a)=-15i+15j-10k[/tex]
Now, [tex](b-a)[/tex] × [tex](c-a)[/tex]
⇒ [tex](5i+25j+5k)[/tex] × [tex](-15i+15j-10k)[/tex]
⇒ [tex]\begin{vmatrix}i&j&k\\5&25&5\\-15&15&-10\end{vmatrix}[/tex]
⇒ [tex]i(-250-75)-j(-50+75)+k(75+375)[/tex]
⇒ [tex]-325i-25j+450k[/tex] -------- (3)
Substitute all values in equation 1,
⇒ [tex][(x-10)i+(y-5)j+(z+50)k].[(-325i-25j+450k)]=0[/tex]
⇒ [tex]-325(x-10)-25(y-5)+450(z+50)=0[/tex]
⇒ [tex]-325x+3250-25y+125+450z+22500=0[/tex]
⇒ [tex]-325x-25y+450z=-22500-3250-125[/tex]
⇒ [tex]-325x-25y+450z=-25875[/tex]
On dividing by 25 and multiply with negative sign on both sides,
⇒ [tex]13x+y-18z=1035[/tex]
Hence we can conclude that the equation of plane containing the points ABC is 13x+y-18z = 1035.
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